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3 years ago

Which applies the power of a power rule properly to simplify this expression?

Mathematics

2 answers:

emmainna [20.7K]3 years ago

6 0

Answer:

Step-by-step explanation:

[7^(-8)]^(-4) =

A power raised to a power: multiply the exponents.

= 7^[(-8)*(4)]

= 7^(32)

Answer: D.

vodomira [7]3 years ago

6 0

Answer:

7^32, the fourth option

Step-by-step explanation:

For the power rule, if you have a number to a power, for example $x^{a}$ and raise it to another power so it becomes, $(x^{a} )^{b}$ , We simplify by multiplying the powers together. So the simplified answer would be $x^{(a*b)}$ .

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How many megabytes of data can a 4.7 gigabyte dvd store?

dedylja [7]

The answer is 4700 megabytes.

1 giga = 1000 mega

1 gigabyte = 1000 megabyte
4.7 gigabyte = x megabyte

1 gb : 1000 mb = 4.7 gb : x
x = 1000 mb * 4.7 gb : 1 gb
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3 years ago

Find the volume of the solid. Round to the nearest tenth if necessary.

Vikki [24]

$V= \frac{4}{3} \pi r^{3} or V=4.1888 r^{3}$

r=4 $r^{3}$ =64*4.1888=268.083

V = 268.083 $mm^{3}$ so go with A.

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3 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

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4 years ago

Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.

masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is $g^{-1}(x) = e^{x}$

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is $g^{-1}(x) = e^{x}$

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y = ln(x)

Taking exponents of both sides, we have

$e^{y} = e^{lnx} \\e^{y} = x$

Replacing x with y, we have

$y = e^{x}$

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is $g^{-1}(x) = e^{x}$

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

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5 0

3 years ago

Smart about slopes help pls

Zielflug [23.3K]

Answer:

Step-by-step explanation:

1. undefined

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3.undefined

6. undefined

thats all i know

7 0

3 years ago

Read 2 more answers